DC Pandey Solutions: Capacitors- 1 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


Introductory Exercise 22.1
1. C
q
V
= = =
- -
[ ]
[ ]
C
AT
[ML T A ]
2 3 1
=
- -
[ ] M L T A
1 2 4 2
    
2. False.
Charge will flow if there is potential
difference between the conductors. It does
not depend on amount of charge present.
3. Consider the charge distribution shown in
figure.
Electric field at point P
    E E E E E
P
= + - -
1 3 2 4
=
-
- + -
- 10
2 2 2
4
2
0 0 0 0
q
A
q
A
q
A
q
A e e e e
But P lies inside conductor
\  E
P
= 0
Þ 10 4 0 - - + - + = q q q q
Þ q = 7 mC             
Hence, the charge distribution is shown in
figure.
Sort-cut Method
Entire charge resides on outer surface of
conductor and will be divided equally on two
outer surfaces.
Hence, if q
1
 and q
2
 be charge on two plates
then.
q q
q q
1 4
1 2
2
3 = =
+
= mC
q
q q
2
1 2
2
7 =
-
= mC        
q
q q
3
2 1
2
7 =
-
= - mC    
Capacitors
22
E
4
E
3
P
E
2
E
1
10–q q –q (q – 4)
1 2 3 4
3mC
7mC –7mC
3mC
q
1
q
2
q
3
q
4
q
2
q
3
q
1
q
4
Page 2


Introductory Exercise 22.1
1. C
q
V
= = =
- -
[ ]
[ ]
C
AT
[ML T A ]
2 3 1
=
- -
[ ] M L T A
1 2 4 2
    
2. False.
Charge will flow if there is potential
difference between the conductors. It does
not depend on amount of charge present.
3. Consider the charge distribution shown in
figure.
Electric field at point P
    E E E E E
P
= + - -
1 3 2 4
=
-
- + -
- 10
2 2 2
4
2
0 0 0 0
q
A
q
A
q
A
q
A e e e e
But P lies inside conductor
\  E
P
= 0
Þ 10 4 0 - - + - + = q q q q
Þ q = 7 mC             
Hence, the charge distribution is shown in
figure.
Sort-cut Method
Entire charge resides on outer surface of
conductor and will be divided equally on two
outer surfaces.
Hence, if q
1
 and q
2
 be charge on two plates
then.
q q
q q
1 4
1 2
2
3 = =
+
= mC
q
q q
2
1 2
2
7 =
-
= mC        
q
q q
3
2 1
2
7 =
-
= - mC    
Capacitors
22
E
4
E
3
P
E
2
E
1
10–q q –q (q – 4)
1 2 3 4
3mC
7mC –7mC
3mC
q
1
q
2
q
3
q
4
q
2
q
3
q
1
q
4
4. Charge distribution is shown in figure.
  q q
q q q
1 4
1 2
2 2
= =
+
= -
q
q q q
2
1 2
2
5
2
=
-
=        
                   q
q q q
3
2 1
2
5
2
=
-
= -
\  Charge on capacitor = Charge on inner
side of positive plate.
q
q
=
5
2
and    C
A
d
=
e
0
\ V
q
C
q d
A
= =
5
2
0
e
Introductory Exercise 22.2
1. All the capacitors are in parallel
q C V
1 1
1 10 10 = = ´ = mC
q C V
2 2
2 10 20 = = ´ = mC
q C V
3 3
3 10 30 = = ´ = mC
2. Potential difference across the plates of
capacitor
               V = 10 V
                q CV = = ´ = 4 10 40 mC
3. In the steady state capacitor behaves as
open circuit.
I =
+
=
30
6 4
3 A
Potential difference across the capacitor,
V I
AB
= ´ = ´ = 4 4 3 12 V
\ Charge on capacitor
q CV
AB
= = ´ = 2 12 24 mC
4. (a) 
1 1 1 1
1
1
2
1 2
C C C
e
= + = +
      C
e
=
2
3
mC
       q C V
e
= = ´ =
2
3
1200 800 mC
(b) V
q
C
1
1
800
1
800 = = = V
   V
q
C
2
2
800
2
400 = = = V
Now, if they are connected in parallel,
Common potential, V
C V C V
C C
=
+
+
æ
è
ç
ç
ö
ø
÷
÷
1 1 2 2
1 2
=
´ ´ ´
+
=
1 800 2 400
1 2
1600
3
 V
q C V
1 1
1600
3
= = mC, q C V
2 2
3200
3
= = mC
5. Common potential
V
C V C V
C C
=
+
+
1 2 2 2
1 2
But V = 20, V
2
0 = , V
1
100 = V, C
1
100 = mC
\ 
100 100 0
400
20
2
2
´ + ´
+
=
C
C
Þ C
2
400 = mC
61 
10V 1mF
C
1
C
3
3mF 2mF
C
2
2W
6W
30V
A B
4W
I
I
V
1
V
2
V
C
1
C
2
C ,V
1 1
C ,V
2 2
Page 3


Introductory Exercise 22.1
1. C
q
V
= = =
- -
[ ]
[ ]
C
AT
[ML T A ]
2 3 1
=
- -
[ ] M L T A
1 2 4 2
    
2. False.
Charge will flow if there is potential
difference between the conductors. It does
not depend on amount of charge present.
3. Consider the charge distribution shown in
figure.
Electric field at point P
    E E E E E
P
= + - -
1 3 2 4
=
-
- + -
- 10
2 2 2
4
2
0 0 0 0
q
A
q
A
q
A
q
A e e e e
But P lies inside conductor
\  E
P
= 0
Þ 10 4 0 - - + - + = q q q q
Þ q = 7 mC             
Hence, the charge distribution is shown in
figure.
Sort-cut Method
Entire charge resides on outer surface of
conductor and will be divided equally on two
outer surfaces.
Hence, if q
1
 and q
2
 be charge on two plates
then.
q q
q q
1 4
1 2
2
3 = =
+
= mC
q
q q
2
1 2
2
7 =
-
= mC        
q
q q
3
2 1
2
7 =
-
= - mC    
Capacitors
22
E
4
E
3
P
E
2
E
1
10–q q –q (q – 4)
1 2 3 4
3mC
7mC –7mC
3mC
q
1
q
2
q
3
q
4
q
2
q
3
q
1
q
4
4. Charge distribution is shown in figure.
  q q
q q q
1 4
1 2
2 2
= =
+
= -
q
q q q
2
1 2
2
5
2
=
-
=        
                   q
q q q
3
2 1
2
5
2
=
-
= -
\  Charge on capacitor = Charge on inner
side of positive plate.
q
q
=
5
2
and    C
A
d
=
e
0
\ V
q
C
q d
A
= =
5
2
0
e
Introductory Exercise 22.2
1. All the capacitors are in parallel
q C V
1 1
1 10 10 = = ´ = mC
q C V
2 2
2 10 20 = = ´ = mC
q C V
3 3
3 10 30 = = ´ = mC
2. Potential difference across the plates of
capacitor
               V = 10 V
                q CV = = ´ = 4 10 40 mC
3. In the steady state capacitor behaves as
open circuit.
I =
+
=
30
6 4
3 A
Potential difference across the capacitor,
V I
AB
= ´ = ´ = 4 4 3 12 V
\ Charge on capacitor
q CV
AB
= = ´ = 2 12 24 mC
4. (a) 
1 1 1 1
1
1
2
1 2
C C C
e
= + = +
      C
e
=
2
3
mC
       q C V
e
= = ´ =
2
3
1200 800 mC
(b) V
q
C
1
1
800
1
800 = = = V
   V
q
C
2
2
800
2
400 = = = V
Now, if they are connected in parallel,
Common potential, V
C V C V
C C
=
+
+
æ
è
ç
ç
ö
ø
÷
÷
1 1 2 2
1 2
=
´ ´ ´
+
=
1 800 2 400
1 2
1600
3
 V
q C V
1 1
1600
3
= = mC, q C V
2 2
3200
3
= = mC
5. Common potential
V
C V C V
C C
=
+
+
1 2 2 2
1 2
But V = 20, V
2
0 = , V
1
100 = V, C
1
100 = mC
\ 
100 100 0
400
20
2
2
´ + ´
+
=
C
C
Þ C
2
400 = mC
61 
10V 1mF
C
1
C
3
3mF 2mF
C
2
2W
6W
30V
A B
4W
I
I
V
1
V
2
V
C
1
C
2
C ,V
1 1
C ,V
2 2
Introductory Exercise 22.3
1. Let q be the final charge on the capacitor, 
work done by battery
W qV =
Energy stored in the capacitor
U qV =
1
2
\ Energy dissipated as heat
H U W qV U = - = =
1
2
2. We have
I I e
t
=
-
0
/ t
        
I
I e
t 0
0
2
=
- /t
 Þ e
t -
=
/t
1
2
      t = = t t ln2 0.693
t =0.693  time constant.
3. Let capacitor C
1
 is initially charged and C
2
 is 
uncharged.
At any instant, let charge on C
2
 be q, charge
on C
1
 at that instant = - q q
0
By Kirchhoff’s voltage law,
( ) q q
C
IR
q
C
0
0
-
- - =
Þ
dq
dt
q q
RC
=
-
0
2
Þ      
dq
q q
dt
RC
q t
0
0 0
2 -
=
ò ò
Þ   
[ln( )]
[ ]
q q
RC
t
q
t 0 0
0
2
2
1 -
-
=
Þ     q
q
e
t
= -
- 0
2
1 ( )
/t
\ At time t,
Charge on C q
q
e
t
1
0
2
1 = = -
-
( )
/ t
Charge on C q q
q
e
t
2 0
0
2
1 = - = +
-
( )
/ t
4. Let q be the charge on capacitor at any
instant t 
By Kirchhoff’s voltage law
q
C
IR E + =
dq
dt
CE q
RC
=
-
        
dq
CE q
dt
RC
q
q t
-
=
ò ò
0
0
Þ q CE e q e
t t
0 0
1 = - +
- -
( )
/ / t t
where, t = RC
5. (a) When the switch is just closed,
Capacitors behave like short circuit.
\ Initial current
I
E
R
i
=
1
(b) After a long time, i.e., in steady state, both
the capacitors behaves open circuit,
I
E
R R
f
=
+
1 3
6. (a) Immediately after closing the switch,
capacitor behaves as short circuit,
 62
V
C
R
R q –q
0
C –(q –q)
1 0
q –q
C
2
I
E
C
I 
+q –q
R
R
1
R
3
S E
R
2
C
2
C
1
Page 4


Introductory Exercise 22.1
1. C
q
V
= = =
- -
[ ]
[ ]
C
AT
[ML T A ]
2 3 1
=
- -
[ ] M L T A
1 2 4 2
    
2. False.
Charge will flow if there is potential
difference between the conductors. It does
not depend on amount of charge present.
3. Consider the charge distribution shown in
figure.
Electric field at point P
    E E E E E
P
= + - -
1 3 2 4
=
-
- + -
- 10
2 2 2
4
2
0 0 0 0
q
A
q
A
q
A
q
A e e e e
But P lies inside conductor
\  E
P
= 0
Þ 10 4 0 - - + - + = q q q q
Þ q = 7 mC             
Hence, the charge distribution is shown in
figure.
Sort-cut Method
Entire charge resides on outer surface of
conductor and will be divided equally on two
outer surfaces.
Hence, if q
1
 and q
2
 be charge on two plates
then.
q q
q q
1 4
1 2
2
3 = =
+
= mC
q
q q
2
1 2
2
7 =
-
= mC        
q
q q
3
2 1
2
7 =
-
= - mC    
Capacitors
22
E
4
E
3
P
E
2
E
1
10–q q –q (q – 4)
1 2 3 4
3mC
7mC –7mC
3mC
q
1
q
2
q
3
q
4
q
2
q
3
q
1
q
4
4. Charge distribution is shown in figure.
  q q
q q q
1 4
1 2
2 2
= =
+
= -
q
q q q
2
1 2
2
5
2
=
-
=        
                   q
q q q
3
2 1
2
5
2
=
-
= -
\  Charge on capacitor = Charge on inner
side of positive plate.
q
q
=
5
2
and    C
A
d
=
e
0
\ V
q
C
q d
A
= =
5
2
0
e
Introductory Exercise 22.2
1. All the capacitors are in parallel
q C V
1 1
1 10 10 = = ´ = mC
q C V
2 2
2 10 20 = = ´ = mC
q C V
3 3
3 10 30 = = ´ = mC
2. Potential difference across the plates of
capacitor
               V = 10 V
                q CV = = ´ = 4 10 40 mC
3. In the steady state capacitor behaves as
open circuit.
I =
+
=
30
6 4
3 A
Potential difference across the capacitor,
V I
AB
= ´ = ´ = 4 4 3 12 V
\ Charge on capacitor
q CV
AB
= = ´ = 2 12 24 mC
4. (a) 
1 1 1 1
1
1
2
1 2
C C C
e
= + = +
      C
e
=
2
3
mC
       q C V
e
= = ´ =
2
3
1200 800 mC
(b) V
q
C
1
1
800
1
800 = = = V
   V
q
C
2
2
800
2
400 = = = V
Now, if they are connected in parallel,
Common potential, V
C V C V
C C
=
+
+
æ
è
ç
ç
ö
ø
÷
÷
1 1 2 2
1 2
=
´ ´ ´
+
=
1 800 2 400
1 2
1600
3
 V
q C V
1 1
1600
3
= = mC, q C V
2 2
3200
3
= = mC
5. Common potential
V
C V C V
C C
=
+
+
1 2 2 2
1 2
But V = 20, V
2
0 = , V
1
100 = V, C
1
100 = mC
\ 
100 100 0
400
20
2
2
´ + ´
+
=
C
C
Þ C
2
400 = mC
61 
10V 1mF
C
1
C
3
3mF 2mF
C
2
2W
6W
30V
A B
4W
I
I
V
1
V
2
V
C
1
C
2
C ,V
1 1
C ,V
2 2
Introductory Exercise 22.3
1. Let q be the final charge on the capacitor, 
work done by battery
W qV =
Energy stored in the capacitor
U qV =
1
2
\ Energy dissipated as heat
H U W qV U = - = =
1
2
2. We have
I I e
t
=
-
0
/ t
        
I
I e
t 0
0
2
=
- /t
 Þ e
t -
=
/t
1
2
      t = = t t ln2 0.693
t =0.693  time constant.
3. Let capacitor C
1
 is initially charged and C
2
 is 
uncharged.
At any instant, let charge on C
2
 be q, charge
on C
1
 at that instant = - q q
0
By Kirchhoff’s voltage law,
( ) q q
C
IR
q
C
0
0
-
- - =
Þ
dq
dt
q q
RC
=
-
0
2
Þ      
dq
q q
dt
RC
q t
0
0 0
2 -
=
ò ò
Þ   
[ln( )]
[ ]
q q
RC
t
q
t 0 0
0
2
2
1 -
-
=
Þ     q
q
e
t
= -
- 0
2
1 ( )
/t
\ At time t,
Charge on C q
q
e
t
1
0
2
1 = = -
-
( )
/ t
Charge on C q q
q
e
t
2 0
0
2
1 = - = +
-
( )
/ t
4. Let q be the charge on capacitor at any
instant t 
By Kirchhoff’s voltage law
q
C
IR E + =
dq
dt
CE q
RC
=
-
        
dq
CE q
dt
RC
q
q t
-
=
ò ò
0
0
Þ q CE e q e
t t
0 0
1 = - +
- -
( )
/ / t t
where, t = RC
5. (a) When the switch is just closed,
Capacitors behave like short circuit.
\ Initial current
I
E
R
i
=
1
(b) After a long time, i.e., in steady state, both
the capacitors behaves open circuit,
I
E
R R
f
=
+
1 3
6. (a) Immediately after closing the switch,
capacitor behaves as short circuit,
 62
V
C
R
R q –q
0
C –(q –q)
1 0
q –q
C
2
I
E
C
I 
+q –q
R
R
1
R
3
S E
R
2
C
2
C
1
\ I
E
R
1
1
= and I
E
R
2
2
=
(b) In the steady state, capacitor behaves as
open circuit,
\ I
E
R
1
1
= , I
2
0 =
(c) Potential difference across the capacitors in 
the steady state,
V E =
\ Energy stored in the capacitor
U CE =
1
2
2
(d) After the switch is open
R R R
e
= +
1 2
          
t = = + RC R R C
3 1 2
( )
AIEEE Corner
Subjective Questions (Level-1)
1. C
A
d
=
e
0
 Þ A
Cd
= =
´ ´
´
-
-
e
0
3
12
1 1 10
10 8.85
        = ´ 1.13 10
8
 m
2
2. C
A
d
1
0 1
=
e
 and C
A
d
2
0 2
=
e
 
If connected in parallel
C C C = +
1 2
 = +
e e
0 1 2 2
A
d
A
d
=
+
=
e e
0 1 2 0
( ) A A
d
A
d
  
where, A A A = + =
1 2
 effective area.
Hence proved.
3. The ar range ment can be con sid ered as the
com bi na tion of three dif fer ent ca pac i tors as
shown in fig ure, where
C
k
A
d
k A
d
1
1 0
1 0 2
2 4
= =
e
e
C
k
A
d
k A
d
2
2 0
2 0 2
2 2 2
= =
e
e
/
C
k
A
d
k A
d
3
3 0
3 0 2
2 2 2
= =
e
e
/
Therefore, the effective capacitance,
C C
C C
C C
= +
+
1
2 3
2 3
       = +
+
æ
è
ç
ç
ö
ø
÷
÷
e
0 1 1 3
2 3
2 2
A
d
k k k
k k
4. (a) Let the spheres A and B carry charges q
and - q respectively,
\ V
q
a
q
d
A
= × -
é
ë
ê
ù
û
ú
1
4
0
pe
   V
q
b
q
d
B
= × - +
é
ë
ê
ù
û
ú
1
4
0
pe
Potential difference between the spheres,
V V V
q
a b d
A B
= - = + -
é
ë
ê
ù
û
ú
4
1 1 2
0
pe
   C
q
V
a b d
= =
+ -
4
1 1 2
0
pe
Hence proved.
63 
R
2
R
1
E
C
C
2
C
3
C
1
d
a b
A B
q –q
A
R
2
I
1
R
1
I
E
B
C
I
2
S
Page 5


Introductory Exercise 22.1
1. C
q
V
= = =
- -
[ ]
[ ]
C
AT
[ML T A ]
2 3 1
=
- -
[ ] M L T A
1 2 4 2
    
2. False.
Charge will flow if there is potential
difference between the conductors. It does
not depend on amount of charge present.
3. Consider the charge distribution shown in
figure.
Electric field at point P
    E E E E E
P
= + - -
1 3 2 4
=
-
- + -
- 10
2 2 2
4
2
0 0 0 0
q
A
q
A
q
A
q
A e e e e
But P lies inside conductor
\  E
P
= 0
Þ 10 4 0 - - + - + = q q q q
Þ q = 7 mC             
Hence, the charge distribution is shown in
figure.
Sort-cut Method
Entire charge resides on outer surface of
conductor and will be divided equally on two
outer surfaces.
Hence, if q
1
 and q
2
 be charge on two plates
then.
q q
q q
1 4
1 2
2
3 = =
+
= mC
q
q q
2
1 2
2
7 =
-
= mC        
q
q q
3
2 1
2
7 =
-
= - mC    
Capacitors
22
E
4
E
3
P
E
2
E
1
10–q q –q (q – 4)
1 2 3 4
3mC
7mC –7mC
3mC
q
1
q
2
q
3
q
4
q
2
q
3
q
1
q
4
4. Charge distribution is shown in figure.
  q q
q q q
1 4
1 2
2 2
= =
+
= -
q
q q q
2
1 2
2
5
2
=
-
=        
                   q
q q q
3
2 1
2
5
2
=
-
= -
\  Charge on capacitor = Charge on inner
side of positive plate.
q
q
=
5
2
and    C
A
d
=
e
0
\ V
q
C
q d
A
= =
5
2
0
e
Introductory Exercise 22.2
1. All the capacitors are in parallel
q C V
1 1
1 10 10 = = ´ = mC
q C V
2 2
2 10 20 = = ´ = mC
q C V
3 3
3 10 30 = = ´ = mC
2. Potential difference across the plates of
capacitor
               V = 10 V
                q CV = = ´ = 4 10 40 mC
3. In the steady state capacitor behaves as
open circuit.
I =
+
=
30
6 4
3 A
Potential difference across the capacitor,
V I
AB
= ´ = ´ = 4 4 3 12 V
\ Charge on capacitor
q CV
AB
= = ´ = 2 12 24 mC
4. (a) 
1 1 1 1
1
1
2
1 2
C C C
e
= + = +
      C
e
=
2
3
mC
       q C V
e
= = ´ =
2
3
1200 800 mC
(b) V
q
C
1
1
800
1
800 = = = V
   V
q
C
2
2
800
2
400 = = = V
Now, if they are connected in parallel,
Common potential, V
C V C V
C C
=
+
+
æ
è
ç
ç
ö
ø
÷
÷
1 1 2 2
1 2
=
´ ´ ´
+
=
1 800 2 400
1 2
1600
3
 V
q C V
1 1
1600
3
= = mC, q C V
2 2
3200
3
= = mC
5. Common potential
V
C V C V
C C
=
+
+
1 2 2 2
1 2
But V = 20, V
2
0 = , V
1
100 = V, C
1
100 = mC
\ 
100 100 0
400
20
2
2
´ + ´
+
=
C
C
Þ C
2
400 = mC
61 
10V 1mF
C
1
C
3
3mF 2mF
C
2
2W
6W
30V
A B
4W
I
I
V
1
V
2
V
C
1
C
2
C ,V
1 1
C ,V
2 2
Introductory Exercise 22.3
1. Let q be the final charge on the capacitor, 
work done by battery
W qV =
Energy stored in the capacitor
U qV =
1
2
\ Energy dissipated as heat
H U W qV U = - = =
1
2
2. We have
I I e
t
=
-
0
/ t
        
I
I e
t 0
0
2
=
- /t
 Þ e
t -
=
/t
1
2
      t = = t t ln2 0.693
t =0.693  time constant.
3. Let capacitor C
1
 is initially charged and C
2
 is 
uncharged.
At any instant, let charge on C
2
 be q, charge
on C
1
 at that instant = - q q
0
By Kirchhoff’s voltage law,
( ) q q
C
IR
q
C
0
0
-
- - =
Þ
dq
dt
q q
RC
=
-
0
2
Þ      
dq
q q
dt
RC
q t
0
0 0
2 -
=
ò ò
Þ   
[ln( )]
[ ]
q q
RC
t
q
t 0 0
0
2
2
1 -
-
=
Þ     q
q
e
t
= -
- 0
2
1 ( )
/t
\ At time t,
Charge on C q
q
e
t
1
0
2
1 = = -
-
( )
/ t
Charge on C q q
q
e
t
2 0
0
2
1 = - = +
-
( )
/ t
4. Let q be the charge on capacitor at any
instant t 
By Kirchhoff’s voltage law
q
C
IR E + =
dq
dt
CE q
RC
=
-
        
dq
CE q
dt
RC
q
q t
-
=
ò ò
0
0
Þ q CE e q e
t t
0 0
1 = - +
- -
( )
/ / t t
where, t = RC
5. (a) When the switch is just closed,
Capacitors behave like short circuit.
\ Initial current
I
E
R
i
=
1
(b) After a long time, i.e., in steady state, both
the capacitors behaves open circuit,
I
E
R R
f
=
+
1 3
6. (a) Immediately after closing the switch,
capacitor behaves as short circuit,
 62
V
C
R
R q –q
0
C –(q –q)
1 0
q –q
C
2
I
E
C
I 
+q –q
R
R
1
R
3
S E
R
2
C
2
C
1
\ I
E
R
1
1
= and I
E
R
2
2
=
(b) In the steady state, capacitor behaves as
open circuit,
\ I
E
R
1
1
= , I
2
0 =
(c) Potential difference across the capacitors in 
the steady state,
V E =
\ Energy stored in the capacitor
U CE =
1
2
2
(d) After the switch is open
R R R
e
= +
1 2
          
t = = + RC R R C
3 1 2
( )
AIEEE Corner
Subjective Questions (Level-1)
1. C
A
d
=
e
0
 Þ A
Cd
= =
´ ´
´
-
-
e
0
3
12
1 1 10
10 8.85
        = ´ 1.13 10
8
 m
2
2. C
A
d
1
0 1
=
e
 and C
A
d
2
0 2
=
e
 
If connected in parallel
C C C = +
1 2
 = +
e e
0 1 2 2
A
d
A
d
=
+
=
e e
0 1 2 0
( ) A A
d
A
d
  
where, A A A = + =
1 2
 effective area.
Hence proved.
3. The ar range ment can be con sid ered as the
com bi na tion of three dif fer ent ca pac i tors as
shown in fig ure, where
C
k
A
d
k A
d
1
1 0
1 0 2
2 4
= =
e
e
C
k
A
d
k A
d
2
2 0
2 0 2
2 2 2
= =
e
e
/
C
k
A
d
k A
d
3
3 0
3 0 2
2 2 2
= =
e
e
/
Therefore, the effective capacitance,
C C
C C
C C
= +
+
1
2 3
2 3
       = +
+
æ
è
ç
ç
ö
ø
÷
÷
e
0 1 1 3
2 3
2 2
A
d
k k k
k k
4. (a) Let the spheres A and B carry charges q
and - q respectively,
\ V
q
a
q
d
A
= × -
é
ë
ê
ù
û
ú
1
4
0
pe
   V
q
b
q
d
B
= × - +
é
ë
ê
ù
û
ú
1
4
0
pe
Potential difference between the spheres,
V V V
q
a b d
A B
= - = + -
é
ë
ê
ù
û
ú
4
1 1 2
0
pe
   C
q
V
a b d
= =
+ -
4
1 1 2
0
pe
Hence proved.
63 
R
2
R
1
E
C
C
2
C
3
C
1
d
a b
A B
q –q
A
R
2
I
1
R
1
I
E
B
C
I
2
S
(b) If d ® ¥
C
a b
ab
a b
=
+
=
×
+
4
1 1
4
0 0
pe pe
If two isolated spheres of radii a and b are
connected in series,
then,
C
C C
C C
¢ =
+
1 2
1 2
where, C a
1 0
4 = pe , C b
2 0
4 = pe
\ C
ab
a b
¢ =
×
+
4
0
pe
\ C C ¢ =
Hence proved.
5. (a) 
(b)
(c)
Let effective capacitance between A and B
C x
AB
=
As the network is infinite,
C C x
PQ AB
= =
Equivalent circuit is shown in figure,
R C
Cx
C x
x
AB
= +
+
=
2
2
Þ 2 2 2
2 2
C Cx Cx Cx x + + = +
Þ x Cx C
2 2
2 0 - - =
On solving, x C = 2 or - C
But x cannot be negative,
Hence,  x C = 2
6. q CV = = ´ = 7.28 C 25 182 m
7. (a) V
q
C
= =
´
´
=
-
-
0.148 10
245 10
604
6
12
 V
(b) C
A
d
=
e
0
 Þ A
Cd
=
e
0
=
´ ´ ´
´
- -
-
245 10 10
10
12 3
12
0.328
8.85
       = ´
-
9.08 m 10
3 2
       =90.8cm
2
(c) s= =
´
´
=
-
-
q
A
0.148
9.08
16.3
10
10
6
3
 mC/m
2
8. (a) E
0
5
10 = ´ 3.20 V/m
    E = ´ 2.50 10
5
 V/m
    k
E
E
= =
´
´
=
0
5
5
10
10
3.20
2.50
1.28
(b)Electric field between the plates of
capacitor is given by
E =
s
e
0
 
Þ s e = = ´ ´ ´
-
0
12 5
10 10 E 8.8 3.20
    = ´
-
2.832 10
6
 C/m
2
 
= 2.832 mC/m
2
  
9. (a) q C V
1 1
4 660 2640 = = ´ = mC
   q C V
2 2
6 660 3960 = = ´ = mC
 64
C
C C
C
C
C
C
C
A B
A B
2C/3
A B
5C/3
A B
2C
A
B
C
C
C
C
C C C C
C C C
A B
A B
C
A B
4C/3
C/3
A
B
C C C C
2C 2C 2C 2C
P
A
B
Q
¥
2C
P
x
Q
C
B
A